Continuous Functions - Thomae's Function ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) b Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 5: Continuous Functions ...

I need help in fully understanding an aspect of Example 5.1.6 (h) ...Example 5.1.6 (h) ... ... reads as follows:
View attachment 7262In the above text from Bartle and Sherbert, we read the following:

" ... ... There are only a finite number of rationals with denominator less than \(\displaystyle n_0 \)in the interval \(\displaystyle ( b - 1, b + 1)\). (Why? ) ... ... Can someone explain to me why the above statement holds true?Help will be appreciated ... ...

Peter*** EDIT ***

... oh ... just realized that if denominator has to be less than \(\displaystyle n_0\) then there can only be, at most, \(\displaystyle n_0 -1\) of these rational numbers ... that is ... a finite number! ... ... ... Is that correct!

Peter

 

[Opalg]

Suppose that $k$ is a positive integer less than $n_0$. The distance between consecutive rational numbers with denominator $k$ is $1/k$. So there at most $k$ such numbers in an interval of length $1$. Therefore there are at most $2k$ such numbers in the interval $(b-1,b+1)$, which has length $2$.

Thus for each of the finitely many numbers $k$ ($1\leqslant k\leqslant n_0$) there are only finitely many rational numbers with denominator $k$ in the interval $(b-1,b+1)$. But a finite union of finite sets is finite. Thus there are only finitely many rational numbers with denominator $\leqslant n_0$ in the interval $(b-1,b+1)$.

 

[Math Amateur]

Suppose that $k$ is a positive integer less than $n_0$. The distance between consecutive rational numbers with denominator $k$ is $1/k$. So there at most $k$ such numbers in an interval of length $1$. Therefore there are at most $2k$ such numbers in the interval $(b-1,b+1)$, which has length $2$.

Thus for each of the finitely many numbers $k$ ($1\leqslant k\leqslant n_0$) there are only finitely many rational numbers with denominator $k$ in the interval $(b-1,b+1)$. But a finite union of finite sets is finite. Thus there are only finitely many rational numbers with denominator $\leqslant n_0$ in the interval $(b-1,b+1)$.

Hmm ... really interesting ...

Required a more thoughtful analysis than my first intuitive reaction ...:( ...

Thanks for the help ... enables me to see what is required in analysis ...

Peter